Dielectric Sphere in a Uniform Field

Drop a glass marble into the gap of a big parallel-plate capacitor and ask an innocent question: what is the field inside the glass? The honest first guess — "the applied field, reduced a bit" — hides a self-consistency loop. The applied field polarizes the glass; the polarization builds bound charge on the surface; the bound charge makes its own field, which changes the field doing the polarizing. The loop looks like it should spiral into a position-dependent mess. For a sphere it does something miraculous instead: it closes in one step, and the interior field comes out perfectly uniform — $3/(\varepsilon_r+2)$ of the applied field, everywhere inside. This problem is the centerpiece of the dielectrics chapter — the place where everything you have built works together as one machine. Linear Dielectrics supplies the material response $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$ ; Boundary Conditions for D and E supplies the seam rules; Separation of Variables in Spherical Coordinates supplies the Legendre machinery; and the Uniformly Polarized Sphere supplies the physical engine — the depolarizing field $-\mathbf{P}/3\varepsilon_0$ — that solves the whole problem a second time in four lines and explains why the answer is uniform. We will earn the $3/(\varepsilon_r+2)$ factor twice, harvest the bound charge, watch the sphere impersonate a point dipole, flip the problem inside out to solve the spherical cavity, and meet the combination $(\varepsilon_r-1)/(\varepsilon_r+2)$ that will follow you through the Clausius–Mossotti relation, Rayleigh scattering, and every effective-medium theory ever built.

Key takeaways

A linear dielectric sphere in a uniform applied field $\mathbf{E}_0$ acquires a uniform interior field $\mathbf{E}_{\mathrm{in}}=\dfrac{3}{\varepsilon_r+2}\,\mathbf{E}_0$ — a privilege of the sphere-and-ellipsoid family only.
From outside, the sphere is exactly a point dipole $\mathbf{p}=4\pi\varepsilon_0R^3\,\dfrac{\varepsilon_r-1}{\varepsilon_r+2}\,\mathbf{E}_0$ on top of the original field — no higher multipoles at any distance.
Two routes agree: the full Legendre boundary-value solve, and a four-line self-consistency loop built on the depolarizing field $-\mathbf{P}/3\varepsilon_0$ of the Uniformly Polarized Sphere.
Limits calibrate everything: $\varepsilon_r\to1$ makes the sphere invisible; $\varepsilon_r\to\infty$ reproduces the conducting sphere, including $\sigma=3\varepsilon_0E_0\cos\theta$ .
The spherical cavity comes from the same master formula: $\mathbf{E}_{\mathrm{cav}}=\dfrac{3\varepsilon_r}{2\varepsilon_r+1}\,\mathbf{E}_0$ — enhanced, never reduced.
Key idea: a uniform field polarizes a sphere uniformly because a uniformly polarized sphere makes a uniform field inside itself — the guess feeds back into itself exactly, and the whole problem collapses to a single dipolar Legendre mode fixed by matching $V$ and $D_\perp$ .

The core insight

A uniform applied field on a sphere produces a uniform polarization because the field of a uniformly polarized sphere is itself uniform inside — so the assumption is self-consistent, and the whole problem collapses to a single $\ell=1$ Legendre mode fixed by matching $V$ and $D_\perp$ .

Where this lives today

The sphere's response factor is a working tool, not a museum piece. Dielectrophoresis sorts living cells in lab-on-chip devices by pushing polarizable spheres with non-uniform fields; the force tracks the real part of exactly the Clausius–Mossotti factor of Eq. (3) (generalized to complex permittivities), so tuning the drive frequency flips cells from attracted to repelled — label-free live/dead sorting on the sign of $(\varepsilon_2-\varepsilon_1)/(\varepsilon_2+2\varepsilon_1)$ . Optical tweezers trap dielectric microspheres by the same gradient force in its optical costume. At the nanoscale the game is engineering the resonance rather than avoiding it: silicon Mie nanoresonators ( $\varepsilon_r\approx12$ at optical frequencies) are the building blocks of dielectric metasurfaces, flat lenses, and structural color; and every effective-medium model of a composite dielectric, from ceramic capacitors to interstellar dust opacity, still starts from Rayleigh's 1892 move of stacking this article's spheres into a lattice.

The physical picture

Put the sphere in the field and watch the first femtoseconds in slow motion. The applied field $\mathbf{E}_0$ stretches every molecule, so a polarization appears pointing along $\mathbf{E}_0$ . Uniform polarization has no volume charge ( $\rho_b=-\nabla\cdot\mathbf{P}=0$ ), but where $\mathbf{P}$ meets the surface it deposits bound charge $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}$ : positive on the downstream hemisphere where the polarization heads out of the material, negative on the upstream one. Those two charged caps are a miniature capacitor wrapped around the sphere, and their field inside points backward — a depolarizing field that fights the applied one.

Figure

The self-consistency loop — applied field, bound-charge caps, and the depolarizing field they feed back

Here is the miracle, said carefully because everything else follows from it. For most shapes the field of the surface bound charge varies from point to point, so the polarization varies, so the bound charge redistributes, and the loop never closes cleanly. But the Uniformly Polarized Sphere proved that uniform $\mathbf{P}$ in a sphere produces a perfectly uniform interior field $-\mathbf{P}/3\varepsilon_0$ . Uniform in, uniform out: guess uniform $\mathbf{P}$ , and the total interior field $\mathbf{E}_0-\mathbf{P}/3\varepsilon_0$ is uniform, and a linear material answers a uniform field with — uniform polarization. The guess reproduces itself. Only the sphere and its ellipsoidal cousins have this property; a dielectric cube in a uniform field has a genuinely nonuniform interior and no closed form. The closed loop is also why the outside world sees something so clean: a uniformly polarized sphere is, exteriorly, a pure point dipole — exactly, every higher multipole identically zero. The sphere's entire disturbance is one dipole moment $\mathbf{p}$ at its center; that single number is what Rayleigh scattering, dielectrophoresis, and effective-medium theory all borrow from this page.

The result, before the work

The answer we are about to earn deserves to be stated cleanly first, because you will use it far more often than you will re-derive it. The interior field is

\mathbf{E}_{\mathrm{in}}=\frac{3}{\varepsilon_r+2}\,\mathbf{E}_0 \qquad (1)

where: $\mathbf{E}_0$ — the uniform applied field (what the plates would make with no sphere present); $\varepsilon_r=\varepsilon/\varepsilon_0$ — the sphere's relative permittivity. Read the dial: $1$ at $\varepsilon_r=1$ ; $3/7\approx0.43$ at $\varepsilon_r=5$ ; $3/82\approx0.037$ for water at low frequency — throttled below $4\%$ . The factor never reaches zero for finite $\varepsilon_r$ : a dielectric shields, but only a conductor shields completely. The polarization follows from the constitutive law applied to the total interior field,

\mathbf{P}=\varepsilon_0\chi_e\,\mathbf{E}_{\mathrm{in}}=3\varepsilon_0\,\frac{\varepsilon_r-1}{\varepsilon_r+2}\,\mathbf{E}_0 \qquad (2)

where: $\chi_e=\varepsilon_r-1$ — the susceptibility delivered by Linear Dielectrics, Susceptibility and Permittivity. The bound surface charge it deposits is $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}=P\cos\theta$ : maximal at the poles, zero on the equator, the $\cos\theta$ caps of the hero figure. Integrating the uniform $\mathbf{P}$ over the volume gives the induced dipole moment,

\mathbf{p}=\tfrac{4}{3}\pi R^3\,\mathbf{P}=4\pi\varepsilon_0R^3\,\frac{\varepsilon_r-1}{\varepsilon_r+2}\,\mathbf{E}_0 \qquad (3)

where: $R$ — the sphere radius. The combination $(\varepsilon_r-1)/(\varepsilon_r+2)$ is the sphere's response dial, running from $0$ (vacuum) to $1$ (conductor). Remember its face: it is the Clausius–Mossotti factor, whose microscopic meaning the Clausius–Mossotti Relation will earn. Pure geometry has already produced the exact combination that later connects $\varepsilon_r$ to molecular polarizability — not a coincidence, but the spherical $1/3$ showing up in both places. The potentials that encode all of this — and that the derivation below actually finds — are

V_{\mathrm{in}}(r,\theta)=-\frac{3}{\varepsilon_r+2}\,E_0\,r\cos\theta,\qquad V_{\mathrm{out}}(r,\theta)=-E_0r\cos\theta+\frac{\varepsilon_r-1}{\varepsilon_r+2}\,\frac{E_0R^3}{r^2}\cos\theta \qquad (4)

where: the polar axis points along $\mathbf{E}_0$ . Inside: a pure $r\cos\theta=z$ potential, i.e. a uniform field. Outside: the applied field plus exactly one dipole term — compare $V_{\mathrm{dip}}=p\cos\theta/4\pi\varepsilon_0r^2$ and you read off (3) at sight.

Derivation AThe full boundary-value solve

We are deriving the interior and exterior potentials of a linear dielectric sphere of radius $R$ and permittivity $\varepsilon=\varepsilon_r\varepsilon_0$ in an asymptotically uniform field $E_0\hat{\mathbf{z}}$ , by the piecewise-Laplace recipe Boundary Conditions for D and E packaged for us. No free charge anywhere, and the hidden-charge theorem of Linear Dielectrics guarantees $\rho_b=0$ in a homogeneous linear region, so $\nabla^2V=0$ holds separately inside and outside; we expand both regions in the axisymmetric Legendre solutions recalled from Separation of Variables in Spherical Coordinates, then stitch the seam: $V$ continuous, $\varepsilon\,\partial V/\partial n$ continuous (since $\sigma_f=0$ ). Watch two things: the far-field condition killing every mode but $\ell=1$ , and the permittivity entering only through the normal-derivative matching — that asymmetry is where the physics lives.

Step 1. Write the general axisymmetric solutions in each region. Regularity at the origin kills the $r^{-(\ell+1)}$ family inside; outside we keep both families:

V_{\mathrm{in}}=\sum_{\ell}A_\ell\,r^{\ell}P_\ell(\cos\theta),\qquad V_{\mathrm{out}}=\sum_{\ell}\Big(C_\ell\,r^{\ell}+\frac{B_\ell}{r^{\ell+1}}\Big)P_\ell(\cos\theta).

Why this form: these are the recalled radial ladders $A_\ell r^\ell+B_\ell r^{-(\ell+1)}$ of Separation of Variables in Spherical Coordinates — we are cashing in owned machinery, not re-deriving it.

Step 2. Impose the far field. As $r\to\infty$ the sphere's disturbance dies away and $V\to-E_0r\cos\theta=-E_0\,r\,P_1(\cos\theta)$ . Matching growing powers term by term forces $C_1=-E_0$ and $C_\ell=0$ for every $\ell\neq1$ .

Step 3. Kill the other modes honestly. For any $\ell\neq1$ the two seam conditions read $A_\ell R^{\ell}=B_\ell R^{-(\ell+1)}$ and $\varepsilon_r\,\ell A_\ell R^{\ell-1}=-(\ell+1)B_\ell R^{-(\ell+2)}$ . Substituting the first into the second gives $\varepsilon_r\,\ell\,A_\ell=-(\ell+1)A_\ell$ , i.e. $\big(\varepsilon_r\ell+\ell+1\big)A_\ell=0$ . The bracket is strictly positive, so $A_\ell=B_\ell=0$ . Why this matters: each mode matches independently (orthogonality), only $\ell=1$ is driven, and every undriven mode's homogeneous system has nonzero determinant. One driving mode in, one responding mode out.

Step 4. Reduce to the dipolar sector and match the potential. With $V_{\mathrm{in}}=A_1r\cos\theta$ and $V_{\mathrm{out}}=-E_0r\cos\theta+B_1\cos\theta/r^2$ , continuity of $V$ at $r=R$ (equivalently of tangential $\mathbf{E}$ ) gives

A_1R=-E_0R+\frac{B_1}{R^2}.

Step 5. Match the displacement. With no free surface charge, $D_\perp=-\varepsilon\,\partial V/\partial r$ is continuous at $r=R$ :

\varepsilon_r A_1=-E_0-\frac{2B_1}{R^3}.

Why $\varepsilon_r$ appears only here: the tangential condition never sees the material — it is pure Faraday-loop geometry — while the normal condition carries $\varepsilon$ because $\mathbf{D}$ , not $\mathbf{E}$ , is what crosses a clean seam unscathed. This is exactly the division of labor Boundary Conditions for D and E promised this problem.

Step 6. Solve the pair. From Step 4, $B_1=R^3(A_1+E_0)$ . Insert into Step 5: $\varepsilon_rA_1=-E_0-2(A_1+E_0)$ , so $(\varepsilon_r+2)A_1=-3E_0$ and

A_1=-\frac{3E_0}{\varepsilon_r+2},\qquad B_1=R^3E_0\Big(1-\frac{3}{\varepsilon_r+2}\Big)=\frac{\varepsilon_r-1}{\varepsilon_r+2}\,R^3E_0.

Step 7. Read off the physics. $\mathbf{E}_{\mathrm{in}}=-\nabla(A_1z)=-A_1\hat{\mathbf{z}}=\dfrac{3}{\varepsilon_r+2}E_0\hat{\mathbf{z}}$ — uniform, because $V_{\mathrm{in}}$ came out purely linear in $z$ . Comparing $B_1\cos\theta/r^2$ with the point-dipole potential gives $p=4\pi\varepsilon_0B_1$ , which is Eq. (3). Why the limits work: at $\varepsilon_r=1$ , $A_1=-E_0$ and $B_1=0$ — the sphere vanishes, as it must when there is no dielectric contrast. As $\varepsilon_r\to\infty$ , $A_1\to0$ (equipotential interior) and $B_1\to R^3E_0$ , reproducing the conducting sphere coefficient for coefficient, including its induced surface charge $\sigma=3\varepsilon_0E_0\cos\theta$ recovered from the $\sigma_b$ harvest below.

The solve is honest but bureaucratic, and it leaves a question the algebra alone cannot answer: why did $V_{\mathrm{in}}$ come out exactly linear? The second route answers that.

Go deeper · Derivation B · 2 minThe self-consistency bootstrap and the geometric series

Derivation BThe self-consistency bootstrap and the geometric series

We are deriving the same interior solution without touching Laplace's equation, by promoting the physical picture to a calculation: assume uniform polarization, use the one fact we own — the Uniformly Polarized Sphere's interior field is exactly $-\mathbf{P}/3\varepsilon_0$ — and demand the constitutive law reproduce the assumed $\mathbf{P}$ . If the loop closes, uniqueness (recalled from the E2 boundary-value theorems) makes this the solution. We close it twice: as a fixed point, and as an iteration whose geometric series shows the feedback physically. Watch the convergence honesty at the end.

Step 1. Superpose the three players. If $\mathbf{P}$ is uniform, the interior field is the applied field plus the depolarizing field:

\mathbf{E}_{\mathrm{in}}=\mathbf{E}_0-\frac{\mathbf{P}}{3\varepsilon_0}.

Why we may write this at all: uniform $\mathbf{P}$ has $\rho_b=0$ and $\sigma_b=P\cos\theta$ , and the Uniformly Polarized Sphere proved that this $\cos\theta$ shell makes the exact uniform interior field $-\mathbf{P}/3\varepsilon_0$ . The $1/3$ is the sphere's depolarizing factor — the single geometric number this whole chapter orbits.

Step 2. Demand material consistency. The medium polarizes in response to the total field, so

\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}_{\mathrm{in}}=\varepsilon_0(\varepsilon_r-1)\Big(\mathbf{E}_0-\frac{\mathbf{P}}{3\varepsilon_0}\Big).

Step 3. Solve the fixed point. Collect $\mathbf{P}$ : $\;\mathbf{P}\big(1+\tfrac{\varepsilon_r-1}{3}\big)=\varepsilon_0(\varepsilon_r-1)\mathbf{E}_0$ , and since $1+\tfrac{\varepsilon_r-1}{3}=\tfrac{\varepsilon_r+2}{3}$ ,

\mathbf{P}=3\varepsilon_0\,\frac{\varepsilon_r-1}{\varepsilon_r+2}\,\mathbf{E}_0,\qquad \mathbf{E}_{\mathrm{in}}=\mathbf{E}_0-\frac{\mathbf{P}}{3\varepsilon_0}=\frac{3}{\varepsilon_r+2}\,\mathbf{E}_0.

The assumed uniform $\mathbf{P}$ produced a uniform $\mathbf{E}_{\mathrm{in}}$ which produced the same uniform $\mathbf{P}$ : the ansatz is exactly self-consistent, and uniqueness upgrades it to the answer.

Step 4. Run the loop as a movie. Start naive: $\mathbf{P}^{(0)}=\varepsilon_0\chi_e\mathbf{E}_0$ . That polarization's caps correct the interior field, giving $\mathbf{P}^{(1)}=\varepsilon_0\chi_e(\mathbf{E}_0-\mathbf{P}^{(0)}/3\varepsilon_0)$ , and so on:

\mathbf{P}^{(n)}=\varepsilon_0\chi_e\mathbf{E}_0\sum_{k=0}^{n}\Big(-\frac{\chi_e}{3}\Big)^{k}\;\xrightarrow{\;n\to\infty\;}\;\frac{\varepsilon_0\chi_e\mathbf{E}_0}{1+\chi_e/3}=3\varepsilon_0\frac{\varepsilon_r-1}{\varepsilon_r+2}\mathbf{E}_0.

For polystyrene ( $\chi_e=1.6$ , ratio $-0.533$ ) the partial sums of $P/\varepsilon_0\chi_eE_0$ run $1,\;0.467,\;0.751,\;0.599,\;0.680,\;0.637,\dots\to0.652=3/(\varepsilon_r+2)$ — overshoot, correct, settle. The factor is an infinite regress of "polarization weakens the field that polarizes," summed. Why the honesty clause: the iteration converges only for $\chi_e<3$ (i.e. $\varepsilon_r<4$ ); for water the movie diverges even though the fixed point of Step 3 is exact for every $\varepsilon_r$ . The series is a picture of the feedback, not the license for the answer — the license is Step 3 plus uniqueness. *This self-consistency result agrees with the boundary-value derivation coefficient for coefficient. As a further check, $\sigma_b=P\cos\theta$ with $P=3\varepsilon_0E_0(\varepsilon_r-1)/(\varepsilon_r+2)\to3\varepsilon_0E_0$ as $\varepsilon_r\to\infty$ , matching the conducting sphere again.*

The exterior world — one dipole, and a tale of two flux maps

Outside, Eq. (4) says the sphere adds exactly one dipole to the world. The computed field map below shows what that does to the lines of $\mathbf{E}$ for $\varepsilon_r=5$ : lines that pass near the sphere are gathered toward it, refract at the seam, cross the interior as sparse straight horizontals — the uniform, throttled $\mathbf{E}_{\mathrm{in}}=\tfrac{3}{7}\mathbf{E}_0$ — and exit symmetrically.

Figure

Field of a dielectric sphere with relative permittivity 5 — computed E-lines, uniform sparse interior

The map hides a subtlety worth owning. Count flux instead of drawing lines: at $\varepsilon_r=5$ the stream function's surface value outside exceeds the interior one by exactly a factor of $5$ ( $\varepsilon_r$ , not a coincidence) — most incoming $\mathbf{E}$ -lines terminate on the negative bound-charge cap rather than crossing. The lines of $\mathbf{D}$ tell the opposite story: with no free charge they never end, and since $D_{\mathrm{in}}=\tfrac{3\varepsilon_r}{\varepsilon_r+2}\varepsilon_0E_0>\varepsilon_0E_0$ , the sphere concentrates displacement flux — $\tfrac{15}{7}$ of the vacuum value at $\varepsilon_r=5$ . $\mathbf{E}$ dilutes inside because bound charge eats its lines; $\mathbf{D}$ concentrates because high permittivity is a low-reluctance path — the two-ledgers audit of Boundary Conditions for D and E, drawn as a picture.

Flip it inside out — the cavity, and one formula to rule both

Nothing in the boundary-value solve cared that vacuum was outside and dielectric inside. Run the identical matching with medium $\varepsilon_1$ outside and $\varepsilon_2$ inside and you get the master formula for a sphere of $\varepsilon_2$ embedded in $\varepsilon_1$ :

\mathbf{E}_{\mathrm{in}}=\frac{3\varepsilon_1}{2\varepsilon_1+\varepsilon_2}\,\mathbf{E}_0,\qquad \mathbf{p}=4\pi\varepsilon_1R^3\,\frac{\varepsilon_2-\varepsilon_1}{\varepsilon_2+2\varepsilon_1}\,\mathbf{E}_0 \qquad (5)

where: $\mathbf{E}_0$ — the uniform field far away in the host medium. Check it: $\varepsilon_1=\varepsilon_0,\ \varepsilon_2=\varepsilon_r\varepsilon_0$ reproduces Eq. (1). Now swap the roles.

Go deeper · Derivation C · 1 minOne master formula for sphere and cavity

Derivation COne master formula for sphere and cavity

We are deriving the field inside a spherical cavity carved out of a large dielectric, as a two-line consequence of the master matching rather than a fresh solve — the payoff of doing the general case once. Strategy: rerun Steps 4–6 of the boundary-value solve with $\varepsilon_1$ outside and $\varepsilon_2$ inside, then specialize. The one thing to watch: which permittivity multiplies which matching equation — the only place the two problems differ.

Step 1. Redo the seam with two media. Potential continuity is unchanged: $A_1R=-E_0R+B_1/R^2$ . Displacement continuity now reads $\varepsilon_2A_1=\varepsilon_1\big(-E_0-2B_1/R^3\big)$ .

Step 2. Solve as before. Eliminating $B_1$ : $(\varepsilon_2+2\varepsilon_1)A_1=-3\varepsilon_1E_0$ , hence Eq. (5). Both earlier results are the $\varepsilon_1=\varepsilon_0$ slice.

Step 3. Specialize to the cavity: $\varepsilon_1=\varepsilon_r\varepsilon_0$ (dielectric host), $\varepsilon_2=\varepsilon_0$ (empty bubble):

\mathbf{E}_{\mathrm{cav}}=\frac{3\varepsilon_r}{2\varepsilon_r+1}\,\mathbf{E}_0.

The factor now runs from $1$ ( $\varepsilon_r=1$ ) upward to $3/2$ ( $\varepsilon_r\to\infty$ ): the cavity field is enhanced, never reduced. Physically the host's polarization charges line the cavity wall with the opposite-sign arrangement — the caps now reinforce the applied field inside the hole. Why the asymmetry with the sphere: the sphere's caps oppose $\mathbf{E}_0$ inside the material that made them; scooping the material out flips which side of the charged shell you stand on. Reduction and enhancement are the same shell seen from opposite sides. *As a consistency check, both factors equal $1$ at $\varepsilon_r=1$ , and each tends to the correct conductor/vacuum limit. Numerically, $\varepsilon_r=5$ gives $15/11=1.364$ (cavity) versus $3/7=0.429$ (sphere), and water ( $\varepsilon_r=80$ ) gives $240/161=1.491$ , within $1\%$ of the ceiling $3/2$ .*

One caution: this is a real, carved cavity — material removed, boundary conditions matched. Local Field and the Lorentz Cavity will introduce a virtual sphere, a bookkeeping surface drawn inside intact material to ask what field one molecule feels; that gives $\mathbf{E}+\mathbf{P}/3\varepsilon_0$ — a different quantity answering a different question. More than one textbook paragraph conflates them; keep the knife-versus-pencil distinction — carve versus draw.

Figure

Sphere versus cavity — the two response dials from the one master formula

Limiting cases

Case	Result	Meaning
$\varepsilon_r\to1$	$\mathbf{E}_{\mathrm{in}}\to\mathbf{E}_0$ , $\mathbf{p}\to0$	No contrast, no response: the sphere is electrically invisible.
$\varepsilon_r\to\infty$	$\mathbf{E}_{\mathrm{in}}\to0$ , $\mathbf{p}\to4\pi\varepsilon_0R^3\mathbf{E}_0$ , $\sigma_b\to3\varepsilon_0E_0\cos\theta$	The conducting sphere of Separation of Variables in Spherical Coordinates, coefficient for coefficient.
weak dielectric, $\chi_e\ll1$	$\mathbf{P}\approx\varepsilon_0\chi_e\mathbf{E}_0\big(1-\tfrac{\chi_e}{3}\big)$	Naive response minus the first depolarization correction — the first term of the bootstrap series.
water, $\varepsilon_r=80$	$\mathbf{E}_{\mathrm{in}}=\tfrac{3}{82}\mathbf{E}_0\approx0.037\,\mathbf{E}_0$	High- $\varepsilon_r$ media nearly expel the field — but only nearly; $3/(\varepsilon_r+2)$ has no zero.
carved cavity	$\mathbf{E}_{\mathrm{cav}}=\tfrac{3\varepsilon_r}{2\varepsilon_r+1}\mathbf{E}_0\in(1,\tfrac32)\,\mathbf{E}_0$	Enhancement, bounded by $3/2$ — the shell seen from the other side.

Common trapThe material responds to the field inside it — not to the field you applied

The most common wrong answer here is $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}_0$ : constitutive law fed the applied field. But $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$ means the total macroscopic field — applied plus the field of every bound charge the polarization itself created. Skipping the feedback loses the entire $3/(\varepsilon_r+2)$ structure and overestimates the polarization by $(\varepsilon_r+2)/3$ — a factor of $27$ for water. If your answer contains $\chi_e\mathbf{E}_0$ with no denominator anywhere, you fed the law the wrong field. (The molecular-scale version of this trap — which field does one molecule feel? — is subtler still, and is the whole subject of Local Field and the Lorentz Cavity.)

Worked examples

Example 1 — a raindrop in a thunderstorm field (confidence builder). A spherical water droplet, $R=1\ \mathrm{mm}$ , $\varepsilon_r=80$ , sits in a storm field $E_0=3\times10^{6}\ \mathrm{V/m}$ (around air's breakdown value). Find the interior field, the polarization, the peak bound-charge density, and the droplet's dipole moment.

Step 1. Interior field from Eq. (1): $E_{\mathrm{in}}=\dfrac{3}{82}\,(3\times10^{6})=1.10\times10^{5}\ \mathrm{V/m}$ . The droplet's interior lives at under $4\%$ of the storm field — water protects its inside remarkably well.

Step 2. Polarization from Eq. (2): $P=3\varepsilon_0\frac{79}{82}E_0=(8.854\times10^{-12})(2.890)(3\times10^{6})=7.68\times10^{-5}\ \mathrm{C/m^2}$ . This is also the peak bound charge: $\sigma_{b,\max}=P=76.8\ \mathrm{\mu C/m^2}$ at the poles, falling as $\cos\theta$ to zero on the equator.

Step 3. Audit the self-consistency. The depolarizing field is $P/3\varepsilon_0=2.89\times10^{6}\ \mathrm{V/m}$ , and indeed $E_0-P/3\varepsilon_0=3.00\times10^{6}-2.89\times10^{6}=1.10\times10^{5}\ \mathrm{V/m}=E_{\mathrm{in}}$ . The caps cancel $96.3\%$ of the applied field.

Step 4. Dipole moment from Eq. (3): $p=4\pi\varepsilon_0R^3\frac{79}{82}E_0=4\pi(8.854\times10^{-12})(10^{-9})(0.963)(3\times10^{6})=\mathbf{3.2\times10^{-13}\ C\cdot m}$ — about $10^{17}$ water-molecule dipoles' worth of aligned moment.

Example 2 — polystyrene sphere, with the seam audited (exam level). A polystyrene sphere, $\varepsilon_r=2.6$ , $R=2\ \mathrm{cm}$ , sits in $E_0=1.0\times10^{5}\ \mathrm{V/m}$ . Find the interior field, the field just outside the pole and just outside the equator, and verify both boundary conditions numerically.

Step 1. Interior: $E_{\mathrm{in}}=\dfrac{3}{4.6}(10^{5})=6.52\times10^{4}\ \mathrm{V/m}$ , uniform, along $\mathbf{E}_0$ . The Clausius–Mossotti factor is $\beta=\dfrac{1.6}{4.6}=0.348$ .

Step 2. Just outside the pole ( $\theta=0$ ) the field is radial with magnitude $E_r^{\mathrm{out}}=E_0(1+2\beta)=1.0\times10^{5}\times1.696=1.696\times10^{5}\ \mathrm{V/m}$ — the applied field plus the dipole's polar doubling.

Step 3. Audit the normal matching there. $D_\perp$ continuity demands $\varepsilon_rE_{\mathrm{in}}=E_r^{\mathrm{out}}$ : $2.6\times6.52\times10^{4}=1.696\times10^{5}$ ✓ — exactly Step 2's number. The $\mathbf{E}$ -jump, $1.04\times10^{5}\ \mathrm{V/m}$ , is carried by $\sigma_{b}=3\varepsilon_0\beta E_0=0.924\ \mathrm{\mu C/m^2}$ : check $\sigma_b/\varepsilon_0=1.04\times10^{5}\ \mathrm{V/m}$ ✓.

Step 4. Just outside the equator the field is tangential: $E_\theta^{\mathrm{out}}=E_0(1-\beta)=6.52\times10^{4}\ \mathrm{V/m}$ — equal to $E_{\mathrm{in}}$ , as tangential continuity demands ( $\sigma_b=0$ on that ring, so the seam is invisible there). Both crossing rules verified with numbers.

Step 5. The dipole moment: $p=4\pi\varepsilon_0R^3\beta E_0=4\pi(8.854\times10^{-12})(8\times10^{-6})(0.348)(10^{5})=\mathbf{3.1\times10^{-11}\ C\cdot m}$ . Bonus flip: an air bubble in polystyrene sits at $E_{\mathrm{cav}}=\frac{3(2.6)}{6.2}E_0=1.26\,E_0$ — a $26\%$ hot spot, which is why voids in high-voltage insulation breed partial discharge.

History

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Terling Place and Cambridge

John William Strutt, third Baron Rayleigh, had the rare habit of solving a small problem so completely that whole fields grew out of the leftovers — and the dielectric sphere was a favorite. In 1871 he used the induced-dipole response of tiny particles — the very $(\varepsilon_r-1)/(\varepsilon_r+2)$ moment of this article — to explain the blue sky via the $\lambda^{-4}$ law that carries his name. In 1892, stacking dielectric spheres into a rectangular lattice, he computed the composite's effective permittivity — founding effective-medium theory with this solution as its only moving part. Nobel laureate (1904) for argon, sometime Cavendish Professor, and an auditor of his own results the way this chapter tries to be: by limits, by symmetry, by experiment.

History

1826

— Poisson solves induced magnetization of spheres and ellipsoids, discovering the shape-dependent demagnetizing (here: depolarizing) factors — the

1/3

enters physics.

1850

— Mossotti connects a dielectric's permittivity to the response of its polarizable particles, the first micro-to-macro bridge through the sphere geometry.

1871

— Rayleigh explains the blue sky with the induced dipole of small dielectric spheres and the fourth-power law.

1873

— Maxwell's Treatise solves the dielectric sphere and layered-sphere problems as set pieces of his new field theory.

1892

— Rayleigh computes the effective permittivity of a lattice of dielectric spheres — effective-medium theory begins.

1904

— Maxwell Garnett turns the single-sphere dipole response into the mixing formula for metal-glass composites still used for nanocomposites today.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §4.4.2 (Example 4.7) · Jackson, Classical Electrodynamics, 3rd ed., §4.4 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 10 · Zangwill, Modern Electrodynamics, §6.5 The key checks are: continuity of $V$ and $D_\perp$ at $r=R$ yields $A_1$ and $B_1$ ; the limits $\varepsilon_r\to1$ (no effect) and $\varepsilon_r\to\infty$ (conducting sphere, $p=4\pi\varepsilon_0R^3E_0$ ) both come out correctly; and the self-consistency route reproduces the same $\mathbf{P}$ via the depolarizing field $-\mathbf{P}/3\varepsilon_0$ .

Builds on: Linear Dielectrics, Susceptibility and Permittivity · Boundary Conditions for D and E · Uniformly Polarized Sphere · Gauss's Law in Dielectrics · The Electric Displacement D

Leads to: Field of a Polarized Object · Induced Dipoles and Atomic Polarizability · Torque and Force on a Dipole in a Field · Polarization Density P

Is the interior field really exactly uniform, or just approximately?

Exactly, in the ideal problem: linear homogeneous isotropic sphere, uniform applied field, no free charge. The solve produces $V_{\mathrm{in}}\propto r\cos\theta=z$ with no higher terms at all, and the self-consistency argument shows why: uniform polarization regenerates itself. Real spheres deviate only insofar as the assumptions do — a non-uniform applied field drives higher multipole responses, and an aspherical droplet picks up corrections of order its eccentricity. The uniformity is a theorem about spheres and ellipsoids, not a small-field approximation.

How can the polarization respond to a field that the polarization itself helped create? It sounds circular.

It is circular — that is the physics, and mathematics handles circularity fine. The constitutive law is local: at every point $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$ with $\mathbf{E}$ the actual total field there. The chicken-and-egg resolves as a fixed point: find the pair $(\mathbf{P},\mathbf{E})$ satisfying the constitutive law and electrostatics simultaneously. The geometric-series derivation shows the loop converging step by step (for $\chi_e<3$ ); the fixed-point algebra works for any $\chi_e$ . Nature does not iterate — the transient settles in femtoseconds and the fixed point is what remains.

Why does only the l = 1 Legendre mode survive when the expansion allows all of them?

Because matching happens mode by mode: Legendre orthogonality means the boundary conditions cannot mix different $\ell$ . The applied field drives only $\ell=1$ ; every other mode's system is homogeneous with determinant $\varepsilon_r\ell+\ell+1>0$ , forcing zero. Tilt the problem — a pear-shaped sphere, a field gradient — and you inject source terms into other modes, which duly light up. One driving harmonic, one responding harmonic.

Does the sphere's field outside really have no higher multipoles even close to the surface?

Really none: the exterior is exactly "uniform field plus ideal dipole" at every $r>R$ , not merely far away. A generic charged blob has an infinite multipole ladder that matters near the source; here the source itself — the $\cos\theta$ bound-charge shell — is a pure $\ell=1$ object. The same purity appeared for the conducting sphere and the Uniformly Polarized Sphere; all three are one statement about $\cos\theta$ shells.

What field does a molecule inside the dielectric actually feel — the E_in of this article?

No — and flagging that is half the point of the next two articles. $\mathbf{E}_{\mathrm{in}}$ is the macroscopic field, a spatial average over many molecules; a single molecule must not count its own contribution, and its discrete neighbors matter. Local Field and the Lorentz Cavity constructs the field that actually polarizes a molecule ( $\mathbf{E}+\mathbf{P}/3\varepsilon_0$ for cubic environments), and the Clausius–Mossotti Relation turns that correction into the bridge between molecular polarizability and $\varepsilon_r$ . This article's sphere is the macroscopic warm-up whose $1/3$ those articles reuse.

Is the depolarizing factor always 1/3, or is that special to the sphere?

Special to the sphere. For a general ellipsoid the interior field is still uniform (the one other shape family with the miracle), but the depolarizing factor splits into three axis-dependent numbers with $N_x+N_y+N_z=1$ ; spherical symmetry forces $1/3$ each way. A long needle along the field has $N\to0$ (the field sails through); a thin slab face-on has $N\to1$ (maximal depolarization, $E_{\mathrm{in}}=E_0/\varepsilon_r$ ) — exactly the slab geometry you handled with Gauss's Law in Dielectrics, a satisfying consistency check.

If the E-field flux into the sphere is throttled, where do the missing field lines go?

Onto the bound charge. Lines of $\mathbf{E}$ end only on charge — free or bound — and the upstream cap's negative $\sigma_b$ terminates most incoming lines at $\varepsilon_r=5$ (the flux bookkeeping gives an exact factor $\varepsilon_r$ between outside and inside); the downstream positive cap launches replacements. $\mathbf{D}$ -lines, blind to bound charge, thread the sphere without ending and get denser inside. Two flux maps, two source rules, one field configuration.

Dielectric Sphere in a Uniform Field

Where this lives today

The physical picture

The result, before the work

The exterior world — one dipole, and a tale of two flux maps

Flip it inside out — the cavity, and one formula to rule both

Limiting cases

Worked examples

History

Going further

Before this

Where this leads